The image of the line frac{x - 1}{3} = frac{y | vedclass

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(C) Let the given line be $L_1: \frac{x - 1}{3} = \frac{y - 3}{1} = \frac{z - 4}{-5} = k$. Any point on $L_1$ is $P(3k + 1, k + 3, -5k + 4)$.To find t

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$\frac{x + 3}{3} = \frac{y - 5}{1} = \frac{z - 2}{-5}$

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(C) Let the given line be $L_1: \frac{x - 1}{3} = \frac{y - 3}{1} = \frac{z - 4}{-5} = k$. Any point on $L_1$ is $P(3k + 1, k + 3, -5k + 4)$.To find the intersection point $B$ of $L_1$ and the plane $2x - y + z + 3 = 0$,substitute the coordinates of $P$ into the plane equation:$2(3k + 1) - (k + 3) + (-5k + 4) + 3 = 0$$6k + 2 - k - 3 - 5k + 4 + 3 = 0$$6 = 0$,which is impossible. This implies the line is parallel to the plane.Let $A(1, 3, 4)$ be a point on the line. The image $A'$ of $A$ in the plane is given by $\frac{x - 1}{2} = \frac{y - 3}{-1} = \frac{z - 4}{1} = -2 \frac{2(1) - 3 + 4 + 3}{2^2 + (-1)^2 + 1^2} = -2 \frac{6}{6} = -2$.So,$x - 1 = -4 \Rightarrow x = -3$,$y - 3 = 2 \Rightarrow y = 5$,$z - 4 = -2 \Rightarrow z = 2$. Thus,$A'(-3, 5, 2)$.The image line passes through $A'(-3, 5, 2)$ and is parallel to the original line,so its direction ratios are $(3, 1, -5)$.The equation of the image line is $\frac{x + 3}{3} = \frac{y - 5}{1} = \frac{z - 2}{-5}$.

The image of the line $\frac{x - 1}{3} = \frac{y - 3}{1} = \frac{z - 4}{-5}$ in the plane $2x - y + z + 3 = 0$ is the line:

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